Bài Giảng: Quy hoạch tuyến tính (Linear Programming)

LOGO
Linear Programming Applications
Operation Research: Deterministic Models

Outline
1. Introduction
2. Marketing Applications
3. Manufacturing Applications
4. Employee Scheduling Applications
5. Transportation Applications
6. Transshipment Applications
7. Ingredient Blending Applications
8. Financial Applications
ISE – Deterministic Optimization Models
2

Introduction
 OR searches for optimal or best solutions
in the presence of resources constraints.
 OR uses mathematical techniques to
model and analyze issues related to
decision making processes
ISE - Production Management
3

Introduction
 Many decisions in management are related with the
best usage resources of organizations.
 Manager makes Decisions in order to satisfy
Objectives, Goals of organizations.
 Resources: Materials, Machines, Man, Money, Time,
Space.
 Linear Programming (LP) is a mathematical method
that helps managers to make decision related with
Resources Allocation. (references about Nobel
laureate: Kantorovich)
 Extensively using computer.
4

Introduction
 Problem: Maximize or Minimize some variables,
usually Profit/ Cost, called Objective function.
 Constraints: are functions show resources limitation
of companies/ organizations. The problem is to find
solution that maximize profits (or minimize lost/cost)
in given constraints.
Form of constraint functions could be:
 Inequality (form  or )
 Equality
 All Objective function and Constraint functions are
linear functions.
5

Marketing Applications
 Linear programming models have been used
in the advertising field as a decision aid in
selecting an effective media mix
 Media selection problems can be
approached with LP from two perspectives
– Maximize audience exposure
– Minimize advertising costs
6

 The Win Big Gambling Club promotes gambling
junkets to the Bahamas
 They have $8,000 per week to spend on advertising
 Their goal is to reach the largest possible high-
potential audience
 Media types and audience figures are shown in the
following table
 They need to place at least five radio spots per week
 No more than $1,800 can be spent on radio
advertising each week
7

MEDIUM
AUDIENCE
REACHED PER AD
COST PER
AD ($)
MAXIMUM ADS
PER WEEK
TV spot (1 minute) 5,000 800 12
Daily newspaper (full-
page ad)
8,500 925 5
Radio spot (30
seconds, prime time)
2,400 290 25
Radio spot (1 minute,
afternoon)
2,800 380 20
 Win Big Gambling Club advertising options
8

Win Big Gambling Club
 The problem formulation is
X1 = number of 1-minute TV spots each week
X2 = number of daily paper ads each week
X3 = number of 30-second radio spots each week
X4 = number of 1-minute radio spots each week
Objective:
Maximize audience coverage = 5,000X1 + 8,500X2 + 2,400X3 + 2,800X4
Subject to X1 ≤ 12 (max TV spots/wk)
X2 ≤ 5 (max newspaper ads/wk)
X3 ≤ 25 (max 30-sec radio spots ads/wk)
X4 ≤ 20 (max 1-min radio spots ads/wk)
800X1 + 925X2 + 290X3 + 380X4 ≤ $8,000 (weekly advertising budget)
X3 + X4 ≥ 5 (min radio spots contracted)
290X3 + 380X4 ≤ $1,800 (max dollars spent on radio)
X1, X2, X3, X4 ≥ 0
9

Win Big Gambling Club
 The problem solution
10

Marketing Research
 Linear programming has also been applied to
marketing research problems and the area of
consumer research
 Statistical pollsters can use LP to help make
strategy decisions
11

Marketing Research
 Management Sciences Associates (MSA) is a marketing research
firm
 MSA determines that it must fulfill several requirements in order
to draw statistically valid conclusions
– Survey at least 2,300 U.S. households
– Survey at least 1,000 households whose heads are 30 years of age
or younger
– Survey at least 600 households whose heads are between 31 and 50
years of age
– Ensure that at least 15% of those surveyed live in a state that
borders on Mexico
– Ensure that no more than 20% of those surveyed who are 51 years
of age or over live in a state that borders on Mexico
12

Marketing Research
 MSA decides that all surveys should be conducted in
person
 It estimates the costs of reaching people in each age
and region category are as follows
COST PER PERSON SURVEYED ($)
REGION AGE ≤ 30 AGE 31-50 AGE ≥ 51
State bordering Mexico $7.50 $6.80 $5.50
State not bordering Mexico $6.90 $7.25 $6.10
13
MSA’s goal is to meet the sampling requirements
at the least possible cost.

Marketing Research
X1 = number of 30 or younger and in a border state
X2 = number of 31-50 and in a border state
X3 = number 51 or older and in a border state
X4 = number 30 or younger and not in a border state
X5 = number of 31-50 and not in a border state
X6 = number 51 or older and not in a border state
 The decision variables are
14

Marketing Research
Objective function
subject to
X1 + X2 + X3 + X4 + X5 + X6 ≥ 2,300 (total households)
X1 + X4 ≥ 1,000 (households 30 or younger)
X2 + X5 ≥ 600 (households 31-50)
X1 + X2 + X3 ≥ 0.15(X1 + X2+ X3 + X4 + X5 + X6) (border states)
X3 ≤ 0.20(X3 + X6) (limit on age group 51+ who can live in
border state)
X1, X2, X3, X4, X5, X6 ≥ 0
Minimize total
interview costs = $7.50X1 + $6.80X2 + $5.50X3
+ $6.90X4 + $7.25X5 + $6.10X6
15

Marketing Research
 The following table summarizes the results of the
MSA analysis
 It will cost MSA $15,166 to conduct this research
REGION AGE ≤ 30 AGE 31-50 AGE ≥ 51
State bordering Mexico 0 600 140
State not bordering Mexico 1,000 0 560
16

Manufacturing Applications
 Production Mix
– LP can be used to plan the optimal mix of
products to manufacture
– Company must meet a myriad of constraints,
ranging from financial concerns to sales demand
to material contracts to union labor demands
– Its primary goal is to generate the largest profit
possible
17

 Fifth Avenue Industries produces four varieties
of ties
– One is expensive all-silk
– One is all-polyester
– Two are polyester and cotton blends
 The table on the below shows the cost and
availability of the three materials used in the
production process
MATERIAL COST PER YARD ($)
MATERIAL AVAILABLE PER
MONTH (YARDS)
Silk 21 800
Polyester 6 3,000
Cotton 9 1,600
Constraints of materials
18

 The firm has contracts with several major department
store chains to supply ties
 Contracts require a minimum number of ties but may
be increased if demand increases
 Fifth Avenue’s goal is to maximize monthly profit
given the following decision variables
19

 Contract data for Fifth Avenue Industries
VARIETY OF TIE
SELLING
PRICE PER
TIE ($)
MONTHLY
CONTRACT
MINIMUM
MONTHLY
DEMAND
MATERIAL
REQUIRED
PER TIE
(YARDS)
MATERIAL
REQUIREMENTS
All silk 6.70 6,000 7,000 0.125 100% silk
All polyester 3.55 10,000 14,000 0.08 100% polyester
Poly-cotton
blend 1
4.31 13,000 16,000 0.10
50% polyester-50%
cotton
Poly-cotton
blend 2
4.81 6,000 8,500 0.10
30% polyester-70%
cotton
VARIETY OF TIE
SELLING
PRICE PER
TIE ($)
MATERIAL
REQUIRED PER
TIE (YARDS)
MATERIAL
COST PER
YARD ($)
COST PER
TIE ($)
PROFIT PER
TIE ($)
All silk $6.70 0.125 $21 $2.62 $4.08
All polyester $3.55 0.08 $6 $0.48 $3.07
Poly-cotton blend
1
$4.31 0.05 $6 $0.30
0.05 $9 $0.45 $3.56
Poly-cotton blend
2
$4.81 0.03 $6 $0.18
0.70 $9 $0.63 $4.00

21
 The decision variables are
X1 = number of all-silk ties produced per month
X2 = number polyester ties
X3 = number of blend 1 poly-cotton ties
X4 = number of blend 2 poly-cotton ties

 The complete Fifth Avenue Industries model
Objective function
Maximize profit = $4.08X1 + $3.07X2 + $3.56X3 + $4.00X4
Subject to 0.125X1 ≤ 800 (yds of silk)
0.08X2 + 0.05X3 + 0.03X4 ≤ 3,000 (yds of polyester)
0.05X3 + 0.07X4 ≤ 1,600 (yds of cotton)
X1 ≥ 6,000 (contract min for silk)
X1 ≤ 7,000 (contract max)
X2 ≥ 10,000 (contract min for all polyester)
X3 ≥ 13,000 (contract min for blend 1)
X4 ≥ 6,000 (contract min for blend 2)
X1, X2, X3, X4 ≥ 0
22

 Solution for Fifth Avenue Industries LP model
23

 Production Scheduling
– Setting a low-cost production schedule over a
period of weeks or months is a difficult and
important management task
– Important factors include labor capacity, inventory
and storage costs, space limitations, product
demand, and labor relations
– When more than one product is produced, the
scheduling process can be quite complex
– The problem resembles the product mix model for
each time period in the future
24

 Greenberg Motors, Inc. manufactures two different
electric motors for sale under contract to Drexel Corp.
 Drexel places orders three times a year for four
months at a time
 Demand varies month to month as shown below
 Greenberg wants to develop its production plan for the
next four months
MODEL JANUARY FEBRUARY MARCH APRIL
GM3A 800 700 1,000 1,100
GM3B 1,000 1,200 1,400 1,400
25

 Production planning at Greenberg must consider
four factors
– Desirability of producing the same number of motors each
month to simplify planning and scheduling
– Necessity to inventory carrying costs down
– Warehouse limitations
– The no-layoff policy
 LP is a useful tool for creating a minimum total
cost schedule the resolves conflicts between these
factors
26

 It costs $10 to produce a GM3A motor and $6 to
produce a GM3B
 Both costs increase by 10% on March 1, thus
 The carrying cost for GM3A motors is $0.18 per
month and the GM3B costs $0.13 per month
 Monthly ending inventory levels are used for the
average inventory level
 Warehouse space capacity: 3300
27

 No worker is ever laid off so Greenberg has a
base employment level of 2,240 labor hours per
month
 By adding temporary workers, available labor
hours can be increased to 2,560 hours per
month
 Each GM3A motor requires 1.3 labor hours and
each GM3B requires 0.9 hours
28

 Double subscripted variables are used in this
problem to denote motor type and month of
production
XA,i = Number of model GM3A motors produced in month i (i =
1, 2, 3, 4 for January – April)
XB,i = Number of model GM3B motors produced in month i
 It costs $10 to produce a GM3A motor and $6 to produce a
GM3B
 Both costs increase by 10% on March 1, thus
Cost of production = $10XA1 + $10XA2 + $11XA3 + 11XA4
+ $6XB1 + $6XB2 + $6.60XB3 + $6.60XB4
29

 We can use the same approach to create the portion
of the objective function dealing with inventory
carrying costs
IA,i = Level of on-hand inventory for GM3A motors at the
end of month i (i = 1, 2, 3, 4 for January – April)
IB,i = Level of on-hand inventory for GM3B motors at the
end of month i
 The carrying cost for GM3A motors is $0.18 per month and
the GM3B costs $0.13 per month
 Monthly ending inventory levels are used for the average
inventory level
Cost of carrying inventory = $0.18IA1 + $0.18IA2 + $0.18IA3 + 0.18IA4
+ $0.13IB1 + $0.13IB2 + $0.13IB3 + $0.13IB4
30

 We combine these two for the objective function
Minimize total cost = $10XA1 + $10XA2 + $11XA3 + 11XA4
+ $6XB1 + $6XB2 + $6.60XB3 + $6.60XB4
+ $0.18IA1 + $0.18IA2 + $0.18IA3 + 0.18IA4
+ $0.13IB1 + $0.13IB2 + $0.13IB3 + $0.13IB4
 End of month inventory is calculated using this
relationship
Inventory
at the end
of this
month
Inventory
at the end
of last
month
Sales to
Drexel this
month
Current
month’s
production
= + –
31

 Greenberg is starting a new four-month production
cycle with a change in design specification that left
no old motors in stock on January 1
 Given January demand for both motors
IA1 = 0 + XA1 – 800
IB1 = 0 + XB1 – 1,000
 Rewritten as January’s constraints
XA1 – IA1 = 800
XB1 – IB1 = 1,000
32

 Constraints for February, March, and April
XA2 + IA1 – IA2 = 700 February GM3A demand
XB2 + IB1 – IB2 = 1,200 February GM3B demand
XA3 + IA2 – IA3 = 1,000 March GM3A demand
XB3 + IB2 – IB3 = 1,400 March GM3B demand
XA4 + IA3 – IA4 = 1,100 April GM3A demand
XB4 + IB3 – IB4 = 1,400 April GM3B demand
 And constraints for April’s ending inventory
IA4 = 450
IB4 = 300
33

 We also need constraints for warehouse space
IA1 + IB1 ≤ 3,300
IA2 + IB2 ≤ 3,300
IA3 + IB3 ≤ 3,300
IA4 + IB4 ≤ 3,300
 No worker is ever laid off so Greenberg has a base employment
level of 2,240 labor hours per month
 By adding temporary workers, available labor hours can be
increased to 2,560 hours per month
 Each GM3A motor requires 1.3 labor hours and each GM3B
requires 0.9 hours
34

 Labor hour constraints
1.3XA1 + 0.9XB1 ≥ 2,240 (January min hrs/month)
1.3XA1 + 0.9XB1 ≤ 2,560 (January max hrs/month)
1.3XA2 + 0.9XB2 ≥ 2,240 (February labor min)
1.3XA2 + 0.9XB2 ≤ 2,560 (February labor max)
1.3XA3 + 0.9XB3 ≥ 2,240 (March labor min)
1.3XA3 + 0.9XB3 ≤ 2,560 (March labor max)
1.3XA4 + 0.9XB4 ≥ 2,240 (April labor min)
1.3XA4 + 0.9XB4 ≤ 2,560 (April labor max)
All variables ≥ 0 Non-negativity constraints
35

 Greenberg Motors solution
PRODUCTION SCHEDULE JANUARY FEBRUARY MARCH APRIL
Units GM3A produced 1,277 1,138 842 792
Units GM3B produced 1,000 1,200 1,400 1,700
Inventory GM3A carried 477 915 758 450
Inventory GM3B carried 0 0 0 300
Labor hours required 2,560 2,560 2,355 2,560
 Total cost for this four month period is $76,301.61
 Complete model has 16 variables and 22 constraints
36

 Assignment Problems
 Involve determining the most efficient way to assign
resources to tasks
 Objective may be to minimize travel times or maximize
assignment effectiveness
 Assignment problems are unique because they have a
coefficient of 0 or 1 associated with each variable in the
LP constraints and the right-hand side of each
constraint is always equal to 1
Employee Scheduling Applications
37

 Ivan and Ivan law firm maintains a large staff of young
attorneys
 Ivan wants to make lawyer-to-client assignments in the
most effective manner
 He identifies four lawyers who could possibly be assigned
new cases
 Each lawyer can handle one new client
 The lawyers have different skills and special interests
 The following table summarizes the lawyers estimated
effectiveness on new cases
38

 Effectiveness ratings
CLIENT’S CASE
LAWYER DIVORCE
CORPORATE
MERGER EMBEZZLEMENT EXHIBITIONISM
Adams 6 2 8 5
Brooks 9 3 5 8
Carter 4 8 3 4
Darwin 6 7 6 4
Let Xij =
1 if attorney i is assigned to case j
0 otherwise
where i = 1, 2, 3, 4 stands for Adams, Brooks, Carter, and
Darwin respectively
j = 1, 2, 3, 4 stands for divorce, merger,
embezzlement, and exhibitionism
39

 The LP formulation is
Maximize effectiveness = 6X11 + 2X12 + 8X13 + 5X14 + 9X21 + 3X22
+ 5X23 + 8X24 + 4X31 + 8X32 + 3X33 + 4X34
+ 6X41 + 7X42 + 6X43 + 4X44
subject to X11 + X21 + X31 + X41 = 1 (divorce case)
X12 + X22 + X32 + X42 = 1 (merger)
X13 + X23 + X33 + X43 = 1 (embezzlement)
X14 + X24 + X34 + X44 = 1 (exhibitionism)
X11 + X12 + X13 + X14 = 1 (Adams)
X21 + X22 + X23 + X24 = 1 (Brook)
X31 + X32 + X33 + X34 = 1 (Carter)
X41 + X42 + X43 + X44 = 1 (Darwin)
40

 Solving Ivan and Ivan’s assignment scheduling LP problem
CLIENT’S CASE
LAWYER DIVORCE
CORPORATE
Adams X
Brooks X
Carter X
Darwin X
CLIENT’S CASE
LAWYER DIVORCE
CORPORATE
Adams 6 2 8 5
Brooks 9 3 5 8
Carter 4 8 3 4
Darwin 6 7 6 4
41

 Labor Planning
 Addresses staffing needs over a particular time
 Especially useful when there is some flexibility in
assigning workers that require overlapping or
interchangeable talents
42

 Hong Kong Bank of Commerce and Industry has
requirements for between 10 and 18 tellers depending on
the time of day
 Lunch time from noon to 2 pm is generally the busiest
 The bank employs 12 full-time tellers but has many part-
time workers available
 Part-time workers must put in exactly four hours per day,
can start anytime between 9 am and 1 pm, and are
inexpensive
 Full-time workers work from 9 am to 5 pm and have 1 hour
for lunch
43

 Labor requirements for Hong Kong Bank of Commerce
and Industry
TIME PERIOD NUMBER OF TELLERS REQUIRED
9 am – 10 am 10
10 am – 11 am 12
11 am – Noon 14
Noon – 1 pm 16
1 pm – 2 pm 18
2 pm – 3 pm 17
3 pm – 4 pm 15
4 pm – 5 pm 10
44

 Part-time hours are limited to a maximum of 50% of the
day’s total requirements
 Part-timers earn $8 per hour on average
 Full-timers earn $100 per day on average
 The bank wants a schedule that will minimize total personnel
costs
 It will release one or more of its part-time tellers if it is
profitable to do so
45

 We let
F = full-time tellers
P1 = part-timers starting at 9 am (leaving at 1 pm)
P4 = part-timers starting at noon (leaving at 4 pm)
P5 = part-timers starting at 1 pm (leaving at 5 pm)
46

subject to
F + P1 ≥ 10 (9 am – 10 am needs)
F + P1 + P2 ≥ 12 (10 am – 11 am needs)
0.5F + P1 + P2 + P3 ≥ 14 (11 am – noon needs)
0.5F + P1 + P2 + P3 + P4 ≥ 16 (noon – 1 pm needs)
F + P2 + P3 + P4 + P5 ≥ 18 (1 pm – 2 pm needs)
F + P3 + P4 + P5 ≥ 17 (2 pm – 3 pm needs)
F + P4 + P5 ≥ 15 (3 pm – 4 pm needs)
F + P5 ≥ 10 (4 pm – 5 pm needs)
F ≤ 12 (12 full-time tellers)
4P1 + 4P2 + 4P3 + 4P4 + 4P5 ≤ 0.50(112) (max 50% part-timers)
P1, P2, P3, P4, P5 ≥ 0
 Objective function
Minimize total daily
personnel cost = $100F + $32(P1 + P2 + P3 + P4 + P5)
47

 There are several alternate optimal schedules Hong
Kong Bank can follow
 F = 10, P2 = 2, P3 = 7, P4 = 5, P1, P5 = 0
 F = 10, P1 = 6, P2 = 1, P3 = 2, P4 = 5, P5 = 0
 The cost of either of these two policies is $1,448 per day
48

Transportation Applications
 Shipping Problem
– The transportation or shipping problem involves
determining the amount of goods or items to be
transported from a number of origins to a number
of destinations
– The objective usually is to minimize total shipping
costs or distances
– This is a specific case of LP and a special
algorithm has been developed to solve it
49

 The Top Speed Bicycle Co. manufactures and markets a line of
10-speed bicycles
 The firm has final assembly plants in two cities where labor
costs are low
 It has three major warehouses near large markets
 The sales requirements for the next year are
– New York – 10,000 bicycles
– Chicago – 8,000 bicycles
– Los Angeles – 15,000 bicycles
 The factory capacities are
– New Orleans – 20,000 bicycles
– Omaha – 15,000 bicycles
50

 The cost of shipping bicycles from the plants to the
warehouses is different for each plant and warehouse
TO
FROM NEW YORK CHICAGO LOS ANGELES
New Orleans $2 $3 $5
Omaha $3 $1 $4
 The company wants to develop a shipping
schedule that will minimize its total annual cost
51

 The double subscript variables will represent the
origin factory and the destination warehouse
Xij = bicycles shipped from factory i to warehouse j
 So
X11 = number of bicycles shipped from New Orleans to New York
X12 = number of bicycles shipped from New Orleans to Chicago
X13 = number of bicycles shipped from New Orleans to Los Angeles
X21 = number of bicycles shipped from Omaha to New York
X22 = number of bicycles shipped from Omaha to Chicago
X23 = number of bicycles shipped from Omaha to Los Angeles
52

 Objective function
Minimize
total
shipping
costs
= 2X11 + 3X12 + 5X13 + 3X21 + 1X22 + 4X23
subject to X11 + X21 = 10,000 (New York demand)
X12 + X22 = 8,000 (Chicago demand)
X13 + X23 = 15,000 (Los Angeles demand)
X11 + X12 + X13 ≤ 20,000 (New Orleans factory supply)
X21 + X22 + X23 ≤ 15,000 (Omaha factory supply)
All variables ≥ 0
53

 Total shipping cost equals $96,000
 Transportation problems are a special case of LP as
the coefficients for every variable in the constraint
equations equal 1
 This situation exists in assignment problems as well as
they are a special case of the transportation problem
 Top Speed Bicycle solution
TO
FROM NEW YORK CHICAGO LOS ANGELES
New Orleans 10,000 0 8,000
Omaha 0 8,000 7,000
54

 Truck Loading Problem
– The truck loading problem involves deciding which
items to load on a truck so as to maximize the
value of a load shipped
– Goodman Shipping has to ship the following six
items
ITEM VALUE ($) WEIGHT (POUNDS)
1 22,500 7,500
2 24,000 7,500
3 8,000 3,000
4 9,500 3,500
5 11,500 4,000
6 9,750 3,500
55

 The objective is to maximize the value of
items loaded into the truck
 The truck has a capacity of 10,000 pounds
 The decision variable is
Xi = proportion of each item i loaded on the truck
56

Maximize
load value
$22,500X1 + $24,000X2 + $8,000X3
+ $9,500X4 + $11,500X5 + $9,750X6
=
 Objective function
subject to
7,500X1 + 7,500X2 + 3,000X3
+ 3,500X4 + 4,000X5 + 3,500X6 ≤ 10,000 lb capacity
X1 ≤ 1
X2 ≤ 1
X3 ≤ 1
X4 ≤ 1
X5 ≤ 1
X6 ≤ 1
X1, X2, X3, X4, X5, X6 ≥ 0
57

 Solution for Goodman Shipping
58

 The Goodman Shipping problem has an interesting issue
 The solution calls for one third of Item 1 to be loaded on
the truck
 What if Item 1 can not be divided into smaller pieces?
 Rounding down leaves unused capacity on the truck and
results in a value of $24,000
 Rounding up is not possible since this would exceed the
capacity of the truck
 Using integer programming, the solution is to load one
unit of Items 3, 4, and 6 for a value of $27,250
59

Transshipment Applications
 The transportation problem is a special case of the
transshipment problem
 When the items are being moved from a source to a
destination through an intermediate point (a
transshipment point), the problem is called a
transshipment problem
60

 Distribution Centers
– Frosty Machines manufactures snowblowers in Toronto and
Detroit
– These are shipped to regional distribution centers in Chicago
and Buffalo
– From there they are shipped to supply houses in New York,
Philadelphia, and St Louis
– Shipping costs vary by location and destination
– Snowblowers can not be shipped directly from the factories
to the supply houses
61

New York City
Philadelphia
St Louis
Destination
Chicago
Buffalo
Transshipment
Point
 Frosty Machines network
Toronto
Detroit
Source
Figure 8.1
62

 Frosty Machines data
TO
FROM CHICAGO BUFFALO
NEW YORK
CITY PHILADELPHIA ST LOUIS SUPPLY
Toronto $4 $7 — — — 800
Detroit $5 $7 — — — 700
Chicago — — $6 $4 $5 —
Buffalo — — $2 $3 $4 —
Demand — — 450 350 300
 Frosty would like to minimize the transportation costs
associated with shipping snowblowers to meet the demands at
the supply centers given the supplies available
63

 A description of the problem would be to minimize cost subject to
1. The number of units shipped from Toronto is not more than 800
2. The number of units shipped from Detroit is not more than 700
3. The number of units shipped to New York is 450
4. The number of units shipped to Philadelphia is 350
5. The number of units shipped to St Louis is 300
6. The number of units shipped out of Chicago is equal to the number
of units shipped into Chicago
7. The number of units shipped out of Buffalo is equal to the number of
units shipped into Buffalo
64

 The decision variables should represent the number of units
shipped from each source to the transshipment points and from
there to the final destinations
T1 = the number of units shipped from Toronto to Chicago
T2 = the number of units shipped from Toronto to Buffalo
D1 = the number of units shipped from Detroit to Chicago
D2 = the number of units shipped from Detroit to Chicago
C1 = the number of units shipped from Chicago to New York
C2 = the number of units shipped from Chicago to Philadelphia
C3 = the number of units shipped from Chicago to St Louis
B1 = the number of units shipped from Buffalo to New York
B2 = the number of units shipped from Buffalo to Philadelphia
B3 = the number of units shipped from Buffalo to St Louis
65

 The linear program is
Minimize cost =
4T1 + 7T2 + 5D1 + 7D2 + 6C1 + 4C2 + 5C3 + 2B1 + 3B2 + 4B3
subject to
T1 + T2 ≤ 800 (supply at Toronto)
D1 + D2 ≤ 700 (supply at Detroit)
C1 + B1 = 450 (demand at New York)
C2 + B2 = 350 (demand at Philadelphia)
C3 + B3 = 300 (demand at St Louis)
T1 + D1 = C1 + C2 + C3 (shipping through Chicago)
T2 + D2 = B1 + B2 + B3 (shipping through Buffalo)
T1, T2, D1, D2, C1, C2, C3, B1, B2, B3 ≥ 0 (nonnegativity)
66

 The solution:
TO
FROM CHICAGO BUFFALO NEW YORK CITY PHILADELPHIA ST LOUIS SUPPLY
Toronto 650 150 — — — 800
Detroit 0 300 — — — 700
Chicago — — 0 350 300 —
Buffalo — — 450 0 0 —
Demand — — 450 350 300
New York City
Philadelphia
St Louis
Destination
Chicago
Buffalo
Transshipment
Point
Toronto
Detroit
Source
650
150
300
450
350
300

Ingredient Blending Applications
 Diet Problems
– One of the earliest LP applications
– Used to determine the most economical diet
for hospital patients
– Also known as the feed mix problem
68

 The Whole Food Nutrition Center uses three bulk grains
to blend a natural cereal
 They advertise the cereal meets the U.S. Recommended
Daily Allowance (USRDA) for four key nutrients
 They want to select the blend that will meet the
requirements at the minimum cost
NUTRIENT USRDA
Protein 3 units
Riboflavin 2 units
Phosphorus 1 unit
Magnesium 0.425 units
69

 We let
XA =pounds of grain A in one 2-ounce serving of cereal
XB =pounds of grain B in one 2-ounce serving of cereal
XC = pounds of grain C in one 2-ounce serving of cereal
GRAIN
COST PER
POUND (CENTS)
PROTEIN
(UNITS/LB)
RIBOFLAVIN
(UNITS/LB)
PHOSPHOROU
S (UNITS/LB)
MAGNESIUM
(UNITS/LB)
A 33 22 16 8 5
B 47 28 14 7 0
C 38 21 25 9 6
 Whole Foods Natural Cereal requirements
70

 The objective function is
Minimize total cost of
mixing a 2-ounce serving = $0.33XA + $0.47XB + $0.38XC
subject to
22XA + 28XB + 21XC ≥ 3 (protein units)
16XA + 14XB + 25XC ≥ 2 (riboflavin units)
8XA + 7XB + 9XC ≥ 1 (phosphorous units)
5XA + 0XB + 6XC ≥ 0.425 (magnesium units)
XA + XB + XC = 0.125 (total mix)
XA, XB, XC ≥ 0
71

 Ingredient Mix and Blending Problems
– Diet and feed mix problems are special cases of a
more general class of problems known as
ingredient or blending problems
– Blending problems arise when decisions must be
made regarding the blending of two or more
resources to produce one or more product
– Resources may contain essential ingredients that
must be blended so that a specified percentage is
in the final mix
72

 The Low Knock Oil Company produces two grades of
cut-rate gasoline for industrial distribution
 The two grades, regular and economy, are created
by blending two different types of crude oil
 The crude oil differs in cost and in its content of
crucial ingredients
CRUDE OIL TYPE INGREDIENT A (%) INGREDIENT B (%) COST/BARREL ($)
X100 35 55 30.00
X220 60 25 34.80
73

 The firm lets
X1 = barrels of crude X100 blended to produce the
refined regular
refined economy
refined regular
refined economy
 The objective function is
Minimize cost = $30X1 + $30X2 + $34.80X3 + $34.80X4
74

 Problem formulation
At least 45% of each barrel of regular must be
ingredient A(X1 + X3) = total amount of crude blended to produce
the refined regular gasoline demand
Thus,
0.45(X1 + X3) = amount of ingredient A required
0.35X1 + 0.60X3 ≥ 0.45X1 + 0.45X3
So
But
0.35X1 + 0.60X3 = amount of ingredient A in refined regular gas
– 0.10X1 + 0.15X3 ≥ 0 (ingredient A in regular constraint)
or
75

 Problem formulation
Minimize cost = 30X1 + 30X2 + 34.80X3 + 34.80X4
subject to X1 + X3 ≥ 25,000
X2 + X4 ≥ 32,000
– 0.10X1 + 0.15X3 ≥ 0
0.05X2 – 0.25X4 ≤ 0
X1, X2, X3, X4 ≥ 0
76

 Solution from QM for Windows
77

Financial Applications
 Portfolio Selection
– Bank, investment funds, and insurance
companies often have to select specific
investments from a variety of alternatives
– The manager’s overall objective is generally to
maximize the potential return on the investment
given a set of legal, policy, or risk restraints
78

 International City Trust (ICT) invests in short-term trade
credits, corporate bonds, gold stocks, and construction
loans
 The board of directors has placed limits on how much
can be invested in each area
INVESTMENT
INTEREST
EARNED (%)
MAXIMUM INVESTMENT
($ MILLIONS)
Trade credit 7 1.0
Corporate bonds 11 2.5
Gold stocks 19 1.5
Construction loans 15 1.8
79

 ICT has $5 million to invest and wants to
accomplish two things
 Maximize the return on investment over the next
six months
 Satisfy the diversification requirements set by the
board
 The board has also decided that at least 55% of the
funds must be invested in gold stocks and
construction loans and no less than 15% be
invested in trade credit
80

 The variables in the model are
X1 = dollars invested in trade credit
X2 = dollars invested in corporate bonds
X3 = dollars invested in gold stocks
X4 = dollars invested in construction loans
81

 Objective function
Maximize
dollars of
interest earned
= 0.07X1 + 0.11X2 + 0.19X3 + 0.15X4
subject to X1 ≤ 1,000,000
X2 ≤ 2,500,000
X3 ≤ 1,500,000
X4 ≤ 1,800,000
X3 + X4 ≥ 0.55(X1 + X2 + X3 + X4)
X1 ≥ 0.15(X1 + X2 + X3 + X4)
X1 + X2 + X3 + X4 ≤ 5,000,000
X1, X2, X3, X4 ≥ 0
82

 The optimal solution to the ICT is to make the
following investments
X1 = $750,000
X2 = $950,000
X3 = $1,500,000
X4 = $1,800,000
 The total interest earned with this plan is
$712,000
83

Bài Giảng: Quy hoạch tuyến tính (Linear Programming)

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Bài Giảng: Quy hoạch tuyến tính (Linear Programming)

Similar to Bài Giảng: Quy hoạch tuyến tính (Linear Programming) (20)

Recently uploaded

Recently uploaded (20)

Bài Giảng: Quy hoạch tuyến tính (Linear Programming)